The Logic of Infons. Yuri Gurevich Itay Neeman

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1 The Logic of Infons Yuri Gurevich Itay Neeman Abstract Infons are pieces of information. In the context of access control the logic of infons is a conservative extension of intuitionistic logic. Distributed Knowledge Authorization Language uses additional unary connectives p said and p implied where p ranges over principals. Here we investigate infon logic and a narrow but useful primal fragment of it. In both cases, we develop model theory and analyze the derivability problem: Does the given query follow from the given hypotheses? Our more involved technical results are on primal infon logic. We construct an algorithm for the multiple derivability problem: Which of the given queries follow from the given hypotheses? Given a bound on the quotation depth of the hypotheses, the algorithm works in linear time. We quickly discuss the significance of this result for access control. Microsoft Research, One Microsoft Way, Redmond, WA 98052, USA Department of Mathematics, UCLA, Los Angeles, CA 90095, USA

2 1 Introduction Infons are pieces of information, e.g. John has the right to read File 13 of the given computer system. The notion of infon is basic in DKAL, Distributed Knowledge Authorization Language, that we have been developing [8, 9, 10]. Quisani 1 : How are infons different from statements? Authors 2 : We never ask whether an infon is true or false. Instead we ask whether it is known to relevant principals. For example, does the owner of File 13 know that John has the right to read the file? Does the administrator of the system know? Does John know? Besides the notion of infon is purely semantical. Q: Did you invent the term? A: No, though for a while we thought we did. The term has been used in situation theory where the name is intended to emphasize the fact that infons are semantic objects, not syntactic representations [7, page 22]. But infons are assigned truth values in situation theory, so our usage of the term is different. One may study the algebra of infons. Order a b if the information in a is a part of that in b. At the bottom of that partial order are uninformative infons carrying no information whatsoever. There is a natural union operation a + b on infons. You know a + b if you know a and you know b. Infon a + b is the least upper bound of a and b. But one has to be careful with the information order because of the omniscience problem well known in epistemic logic. The partial order is intractable. Indeed, valid logic statements are uninformative. In [8] we used a rule-based feasible approximation of the true information order. The right tool to deal with infons is logic. The addition operation a + b can be thought as conjunction a b. And the implication connective is natural. If you know a and you know a b then you know b. In terms of the information order, a b is the least solution x of the inequality a + x b. Q: Give an example of implication. A: You have the right to fish in this river if you have an appropriate licence from the state. Implication allowed us to simplify DKAL [10]. For example principal p is trusted on saying infon x, used to be a primitive notion in DKAL subject to a stipulation x (p said x) + (p trusted on saying x). Now we define p trusted on saying x as (p said x) x. The stipulation follows. It turns out that infon logic is a conservative extension of (disjunction-free, propositional) 1 Quisani is an inquisitive friend of ours. 2 Speaking one at a time. 2

3 intuitionistic logic. In addition to conjunction and implication, infon logic has two unary connectives p said and p implied for any principal p. The number of principals is unbounded. Q: How is infon logic related to authorization? A: In DKAL, a principal uses infon logic to derive consequences from (i) his own knowledge assertions and (ii) the communications from other principals. Q: How are said and implied different? A: A principal p may just say x to q. Then (if the communication reaches q and q accepts it but let us ignore all that) q learns infon p said x. However p may condition his saying x on q knowing y. Then q learns infon y (p implied x). Q: Give me an example. A: Suppose that Alice and Bob work in the same company, and Charlie is an outside visitor. If Alice tells Bob that Charlie can read the latest company letter then Bob learns this: Alice said Charlie can read the letter. But if Alice tells Bob that Charlie can read the letter provided that Charlie signed the non-disclosure agreement (NDA) then Bob learns this: if Charlie signed the NDA then Alice implied that Charlie can read the letter. Q: What does this buy you? Suppose that Bob knows the proviso that Charlie signed the NDA. Then he learns that Alice implied that Charlie can read the letter. Shouldn t he learn that Alice said that Charlie can read the letter? A: If one is not careful, a proviso can be used for undesired delegation and even a probing attack [10, 7]. To get from knowing p said x to knowing x, q needs to trust p on saying x. To get from knowing p implied x to knowing x, q needs to trust p on implying x which is a stronger condition. The derivability problem (whether given formulas Γ entail another given formula x) for intuitionistic logic may seem to be intractable. Even the termination of the proof search does not seem to be guaranteed. But there are intuitionistic calculi with the subformula property: if Γ entails x then there is a derivation of x from Γ that uses only subformulas of formulas Γ {x}. This helps to establish that the problem is solvable in polynomial space. The problem is in fact polynomial space complete [16]. Starting with a known sequent calculus for intuitionistic logic which is sound and complete with respect to the standard Kripke semantics and which has the subformula property, we extend the calculus and semantics to infon logic and prove that the extended calculus is sound and complete and has the subformula property. The derivability problem remains polynomial space complete in the 3

4 worst case. This does not make infon logic useless. It works in great many practical cases. Typical cases are far from the worst ones. Further, we identified a narrow primal fragment of infon logic that is surprisingly expressive. For the sake of contrast, the original infon logic may be called full. We modify the sequent calculus for full infon logic to fit primal infon logic. The new calculus has a version of subformula property; instead of subformulas we speak of relatives of theirs which we call local (to the given hypotheses and query) formulas. The new calculus is sound and complete with respect to Kripke semantics adjusted in a simple and natural way. The definition of when an implication x y holds in a world w becomes non-deterministic and that s it. Q: What about primal intuitionistic logic? I mean the intersection of primal infon logic with intuitionistic logic. It s got to be known. Intuitionistic logic has been researched so thoroughly for such a long time. A: Well, the only references of relevance that we know are proof-theoretic papers [1] and [2]. Their semi-implication is our primal implication. But even the two papers have not been published yet; we learned of them when we presented infon logic at Tel Aviv University. Primal intuitionistic logic is weak indeed. That may explain the lack of attention. The more involved technical results of this paper are related to the derivability problem for primal infon logic. Definition 1.1. The multiple derivability problem MD(L) for a logic L is to compute, given formulas x 1,..., x m (the hypotheses) and y 1,..., y n (the queries), which of the queries are derivable from the hypotheses. We construct an algorithm that solves the multiple derivability problem for primal infon logic. We stratify primal logic according to the quotation depth of infons. For example, both infons Alice said that Bob implied that John has the right to read File 13 and Bob said that John has the right to read File 13 belong to stratum 2 but only the second belongs to stratum 1. At each stratum, our algorithm works in linear time. (Our computation model is the usual random access machine, as in [6].) In applications quotation depth tends to be small. Many authorization languages do not even allow nesting the pronouncements of principals. This applies to all Datalog-based authorization languages that we know. Q: The lowest stratum where pronouncements is nested is stratum 2. Is stratum 2 of primal logic of any use for authorization purposes? A: Very much so. In our experience, stratum 2 suffices for most purposes. Q: Is there any evidence beyond your own experience? A: An authorization language SecPAL [3] expresses many important access control scenarios. A rather natural translation of SecPAL to DKAL uses only stratum 2 of primal infon logic. 4

5 This paper is self-contained. We do not presume that the reader is familiar with DKAL or intuitionistic logic. Acknowledgments Conversations with Nikolaj Bjørner, Andreas Blass and Grigori Mints have been most helpful. Last-minute remarks of Sergio Mera were useful as well. 2 Sequent calculi for full and primal infon logic In DKAL, infons are expressed by means of terms, and formulas have the form principal knows infon. Primitive terms have the form t 0 A(t 1,..., A k ) where A is an attribute name, each t i is a term in the sense of first-order logic, and t 0 is of type Principal. Compound terms are built from primitive ones by means of functions and of type Infon Infon Infon and functions said and implied of type Principal Infon Infon. This paper is devoted to infon logic. (Readers interested primarily in DKAL and not so much in infon logic per se may want to go directly to [10].) Here we treat infon terms as propositional formulas. We use symbol to represent an uninformative infon. We presume an infinite vocabulary of infon variables and another infinite vocabulary of (the names of) principals. Formulas are built from infon variables and the infon constant by the following means. Conjunction. If x, y are formulas then so is x y. Implication. If x, y are formulas then so is x y. Two unary connectives p said and p implied for every principal p. If x is a formula then so are p said x and p implied x. In the sequel, formulas are by default these formulas. Formulas that do not involve the unary connectives will be called quotation-free. Some of our algorithms take formulas or sequences of formulas as inputs. In this connection, we presume that the syntax of formulas is such that an occurrence of a subformula y in a formula x is uniquely defined by the position of the first symbol of y in x. 2.1 Sequent calculus for full infon logic Our sequent calculus SCF for full infon logic is essentially an extension of the disjunction-free version of the intuitionistic propositional system NJp [14, 2.2]. A sequent has the form Γ x where x is a formula and Γ is a set of formulas written as a sequence. Here are the axioms and rules of inference. 5

6 Axioms ( ) (x2x) x x Inference rules (Premise Inflation) Γ y Γ, x y ( E) ( I) ( E) ( I) (Said) (Implied) Γ x y Γ x Γ x Γ y Γ x y Γ x Γ, x y Γ x y Γ x y Γ y Γ x y Γ y y q said q said y ( 1 2 ) y (q said 1 ) (q implied 2 ) q implied y Here E and I allude to elimination and introduction respectively. If = {x 1,..., x n } then q said is the set {(q said x i ) : i = 1,..., n} and q implied is the set {(q implied x i ) : i = 1,..., n}. Corollary 2.1. q said x q implied x. Proof Apply rule (Implied) to x x. Corollary 2.2 (Transitivity). If Γ x and Γ, x y then Γ y. Proof By ( I), we have Γ (x y). It remains to apply ( E). Q: I guess is the propositional constant usually called true. A: This is a reasonable point of view as far as this paper is concerned. Q: Is there any semantical difference between and true? 6

7 A: Yes, is a piece of information known to all principals. It is natural to assume that this infon is true but we don t have to lump all true infons together. Some true infons may be informative. Q: Rule (Implied) is too complicated. Turn Corollary 2.1 into an axiom and then replace (Implied) with a simpler rule y q implied q implied y. A: Calculus SCF has the subformula property: If a sequent Γ x is provable then it has a proof that uses only subformulas of the sequent formulas; see Theorem 3.2 below. The modified system loses the subformula property. For example any proof of sequent s = [q said r said x q implied r implied x]. involves additional formulas. 2.2 Sequent calculus for primal infon logic We introduce primal infon logic by means of a sequent calculus SCP obtained from the calculus SCF for full infon logic by replacing the implicationintroduction rule ( I) with two rules ( IW) (Trans) Γ y Γ (x y) Γ x, Γ, x y Γ y Rule ( IW) is a weaker (hence the W ) implication-introduction rule. Rule (Trans) reflects Corollary 2.2. Lemma 2.3. In either calculus, if Γ {x 1,..., x n } y and Γ x 1,..., Γ x n then Γ y. Proof Induction on n. If n = 0, the lemma is trivial. Suppose that n > 0 and the lemma has been proved for n 1. Let Γ = Γ {x 1,..., x n 1 }. Use (Premise Inflation) to derive Γ x n from the last premise. The first premise is Γ, x n y. Now use (Trans) and then the induction hypothesis. Q: Rule ( IW) is ridiculous. Why do you need the implication if you already have the conclusion? A: A principal may know the conclusion y but may be willing to share only an implication x y with another principal. Besides the implication x y may be needed for derivation as it may occur as the premise of another implication. 7

8 3 Semantics, soundness and completeness, subformula property 3.1 Semantics for full infon logic Recall that a quasi-order is a binary relation that is reflexive and transitive. A subset U of a quasi-ordered set (W, ) is a cone if u U and u v imply v U for all u, v in W. A Kripke structure for propositional intuitionistic logic [12] can be defined as a triple (W,, C) where K1 W is a nonempty set whose elements are traditionally called worlds, K2 is a quasi-order of W, and K3 C assigns a cone to every infon variable. By induction, every quotation-free formula z is assigned a cone C(z) of worlds: K4. C( ) = W. K5. C(x y) = C(x) C(y). K6. C(x y) = {u : C(x) {v : v u} C(y)}. Here plays the traditional role of propositional constant true. It is easy to check, by induction on formula z, that every C(z) is indeed a cone. We extend this definition to accommodate full infon logic. A Kripke structure for full infon logic is a quintuple (W,, C, S, I) where W,, C are as above and where S and I assign binary relations S q and I q over W respectively to every principal q in such a way that the following two requirements are satisfied. K7. I q S q. K8. If u w and w S q v then u S q v, and the same for I q. The cone map C is extended to all formulas by means of clauses K4 K6 and the following two clauses. K9. C(q said x) = {u : {v : u S q v} C(x) }. K10. C(q implied x) = {u : {v : u I q v} C(x) }. If u C(z), we say that z holds in u and that u models z, and we write u = z. Again, it is easy to check, by induction on formula z, that every C(z) is indeed a cone. We consider here only the case when z = (q said x). Suppose that u C(z) and u w. By K9, {v : u S q v} C(z) and we need to show that {v : w S q v} C(z). This follows from K8. The cone map C extends naturally to sets of formulas and to sequents: C(Γ) = x Γ C(x), 8

9 C(Γ y) = {u : C(Γ) {v : v u} C(y)}. The Kripke structure itself models a sequent s = [Γ y] if C(s) = W which is equivalent to C(Γ) C(y). A sequent s is valid if every Kripke structure models s. Corollary 3.1. C(q said x) C(q implied x) Proof By K9 and K10, it suffices to show that, for every u, {v : u I q v} {v : u S q v}. This is exactly K Semantics for primal infon logic The definition of Kripke structures for primal infon logic is similar to that for full infon logic except that the deterministic requirement K4 is replaced with the following non-deterministic requirement. K6W. C(x y) is an arbitrary cone subject to constraint C(y) C(x y) {u : C(x) {v : v u} C(y)}. 3.3 Soundness and completeness Theorem 3.2. In the case of either infon logic, full or primal, the following claims are equivalent for any sequent s. 1. s is provable. 2. s is valid. 3. Every finite Kripke structure models s. 4. There is a proof of s that uses only subformulas of s. Proof We deal with both logics at once. We prove that (1) = (2) = (3) = (4) = (1). Implications (4) = (1), and (2) = (3) are obvious. (1) = (2). Let K = (W,, C, S, I) be an arbitrary Kripke structure. By induction on the given derivation of sequent s we show that K models s. If s is the ( ) axiom, use K4. If s is an (x2x) axiom, the desired C(x) C(x) is obvious. Suppose that s is not an axiom. Let u be a world in the cone of the premise of s. We need to show that u models the conclusion of s. Consider the last step in the given derivation of s. Several cases arise. The case of (Premise Inflation) is obvious. The cases ( E) and ( I) are obvious as well; just use K5. ( E). By the induction hypothesis, u models x as well as x y. By K6W (use the second inclusion of the constraint), u models y. 9

10 ( I). This case is relevant for the full but not primal infon logic. By K6, it suffices to check that v = x implies v = y for all v u. Suppose v = x. By the induction hypothesis, v = y. ( IW). This case is relevant for primal infon logic. By the induction hypothesis, u models y. By K6W (use the first inclusion of the constraint), u models x y. (Trans). This case is relevant for primal infon logic. By the choice of u, we have u C(Γ). By the induction hypothesis applied to the first premise, u C(x). Now we apply the induction hypothesis to the second premise: u C(y). (S) and (I). These two cases are similar. We consider only (I). By the choice of u, we have u C(q said 1 ) C(q implied 2 ). By Corollary 3.1, u C(q implied ( 1 2 )) which, by K10, is equivalent to this: v C( 1 2 ) for all v with u I q v. For any such v, by the induction hypothesis, v C(y). By K10, u C(q implied y). (3) = (4). Assuming that (4) fails, we construct a finite model K = (W,, C, S, I) for sequent s. Call a formula native if it is a subformula of s. A native theory is any set u of native formulas closed under native deduction in the following sense: u contains every native formula x such that sequent u x is provable using only native formulas. The quasi-order (W, ) of K is the set of native theories ordered by inclusion. Set u I q v true if v contains every formula x such that (q implied x) belongs to u or (q said x) belongs to u. Set u S q v true if v contains every formula x such that q said x belongs to u. Requirements K1 K2 and K7 K8 are obviously satisfied. It remains to define the cone map C. In the case of full infon logic, we could have defined C on the variables and then used clauses K4 K6 and K9 K10 to extend C to compound formulas. For the uniformity of the proof, we choose a different route. If z is a native formula or define C(z) = {u : z u}, so that u = z if and only if z u. Clearly requirements K3 and K4 are satisfied. Now use clauses K5 K6 and K9 K10 to extend C to the remaining formulas composed from native formulas and by means of connectives,, q said, q implied. To complete the definition of K, we check that requirements K5, K6 (resp. K6W) and K9 K10 on C(z) are satisfied for any native formula z. This is done by induction on z. The base of induction, when z is a variable, is trivial. The induction step splits into several cases. Case K5. Using the definition of C on native formulas and the fact that every world is closed under native deduction, we have u C(x y) (x y) u x u y u u C(x) C(y). Case K6. First we prove that C(x y) {u : C(x) {v : v u} C(y)}. This part is relevant to both infon logics. Suppose that u contains x y. If a native theory v u also contains x then it contains y. Second we prove that {u : C(x) {v : v u} C(y)} C(x y). This part is relevant only to full infon logic. Pick an arbitrary world u such 10

11 that C(x) {v : v u} C(y). We claim that sequent u, x y is provable. Otherwise, there is a native theory v that includes u, contains x but does not contain y which contradicts the choice of u. By ( I) with Γ = u, we have that u contains x y. Case K6W. This case is relevant for primal infon logic. The right inclusion has been already proven. To prove the left inclusion, suppose that u C(y), so that u y is provable. By ( IW), u (x y) is provable, and so u C(x y). Cases K9 and K10. The two cases are similar; we consider only case K9. Consider an arbitrary world u. First suppose that u C(q said x), that is u contains (q said x). We need to prove that any v with u S q v contains x. This follows from the definition of S q. Second suppose that {v : u S q v} C(x), that is every native theory v with u S q v contains x. Let be the set of formulas y such that u contains (q said y), and let be the least native theory that includes. By the definition of S q, we have u S q. Then contains x, so that the sequent x is provable. By rule (Said), sequent u (q said x) is provable, and so u contains q said x. Thus K is a legitimate Kripke structure. Finally let s be Γ x, and let Γ be the least native theory. By the assumption that (4) fails, Γ does not contain x. It follows that s fails in K. 4 Full infon logic: complexity Theorem 4.1. The validity problem (whether a given formula is valid) for full infon logic is polynomial-space complete. The rest of this section is devoted to proving the theorem. The quotation-free fragment of our sequent calculus for full infon logic is a calculus for a fragment of propositional intuitionistic logic whose validity problem is pspace (that is polynomial space) hard [16]. Hence our problem is pspace hard. It remains to show that the validity problem for full infon logic is pspace. We use the following idea that goes back to Ladner [13] who proved that the validity problem for some modal logics is pspace. Instead of checking validity, check whether the given formula can be refuted in a tree-like model structure, which may be exponentially large but where the branches have only polynomial length and thus can be constructed one branch at a time. The idea was developed in a variety of publications, in particular by Halpern and Moses [11] and by Schröder and Pattinson [15]. The latter paper is quite recent, has a survey of related literature and will delight the fans of category theory. Instead of constructing a tree-like model structure and examining it branch by branch, we use games; this helps to avoid bookkeeping. Recall that pspace equals alternating polynomial time [5]. We show that the unprovability problem for the sequent calculus SCF for full infon logic, whether a given sequent is unprovable, is solvable in alternating polynomial time. 11

12 We start with a few auxiliary definitions. A sequent s = [Γ ϕ] self-refuting if Γ is closed under s-native deduction but does not contain ϕ. A formula is potent if it has the form x y or p said x or p implied x. Define quotation depth in the obvious way: QD(z) = 0 if z is a variable or, QD(x y) = QD(x y) = max{qd(x), QD(y)}, QD(p said x) = QD(p implied x) = 1 + QD(x), QD(Γ z) = max{qd(x) : x Γ x = z}. The game Given a sequent s 0 = [Γ 0 ϕ 0 ], we define a game G(s 0 ) between two players: Refuter, who intends to refute s, and Challenger. Let n be the length of s 0 and d the quotation depth QD(s 0 ). Notice that the number of s 0 -native formulas is n. Phase 0. Refuter starts the game by guessing a sequent s 1 = [Γ 1 ϕ 1 ] such that Γ 1 Γ 0 and ϕ 1 = ϕ 0 ; Refuter claims that s 1 is self-refuting. Winning condition. For any k > 0, the state of the game after phase k 1 (unless the game terminated earlier) is given by a sequent s k = [Γ k ϕ k ]. The game continues further provided the following conditions are satisfied. R1 Γ k, and ϕ / Γ k, and every formula in Γ k is s k 1 -native. R2 If x y is in Γ k then both x and y are in Γ k. R3 If x, y are in Γ k and x y is s k -native then x y is in Γ k. R4 If x is in Γ k and x y is in Γ k then y is in Γ k. C1 There is an s k native potent formula outside of Γ k. Otherwise the game terminates. If at least one of the conditions R1 R4 fails then Challenger wins. If conditions R1 R4 hold but condition C1 fails, then Refuter wins. Phase number k > 0. In state given by sequent s k = [Γ k, ϕ k ], Challenger chooses a potent s k -native formula z outside of Γ k. Case z = (x y). Refuter guesses a sequent s k+1 = [Γ k+1 y] with Γ k+1 Γ k {x}. Case z = (p said y). Let be the set of formulas x such that formula p said x belongs to Γ k. Refuter guesses a sequent s k+1 = [Γ k+1 y] where Γ k+1 consists of formulas native to {y} and Γ k+1. Case z = (p implied y). Let be the set of formulas x such that either (p said x) or (p implied x) belongs to Γ k. Refuter guesses a sequent s k+1 = [Γ k+1 y] where Γ k+1 consists of formulas native to {y} and Γ k+1. 12

13 That completes the description of the game. Termination The game terminates in at most 1 + (d + 1)n phases. Indeed, at every phase k where Challenger chooses a said-formula or implied-formula, we have QD(s k+1 ) < QD(s k ). Thus there can be at most d such cases. And every stretch of phases where Challenger chooses implications can contain at most n phases as at each such phase the set of hypotheses grows but there can only be n hypotheses as there are only n s 0 -native formulas. The only exception is the very first stretch because the set of hypotheses may not grow during the very first phase. Lemma If s 0 is refutable then Refuter has a winning strategy in G(s 0 ). 2. If s 0 is valid then Challenger has a winning strategy in G(s 0 ). Proof 1. Assuming that s 0 = [Γ 0 ϕ 0 ] is refutable, we construct a winning strategy for Refuter. At phase 0, Refuter chooses Γ 1 to be the least s 0 -native theory that includes Γ 0, and he chooses ϕ 1 = ϕ 0 so that sequent s 1 = [Γ 1 ϕ 1 ] is self-refuting. Suppose that k > 0, phase k 1 has been executed, the current sequent s k is self-refuting, the game continues, and Challenger chose a formula z. We show that Refuter can reply with a self-refuting sequent s k+1 = Γ k+1 ϕ k+1 whose form depends on the form of z. But first notice that s k is not valid because it is self-refuting and condition R1 holds. Let K be the counter-model constructed in the proof of Theorem 3.2 with s k playing the role of s. Since s k is self-refuting, Γ k is an s k -native theory and thus a world in K. Case z = (x y). Since (x y) does not belong to s k -native theory Γ k, there is a world Γ k+1 Γ k in K that contains x but not y. The desired s k = [Γ k+1 y]. Case z = (p said y). Let be as in the definition of phase k > 0. Taking into account the rule (Said) of calculus SCF for full infon logic, we see that sequent t = y is unprovable. The desired Γ k+1 is the least t-native theory that includes and the desired ϕ k+1 = y. Case z = (p implied y) is similar to the previous one. 2. Suppose that s 0 is valid and thus provable in calculus SCF for full infon logic. We construct a winning strategy for Challenger. Suppose that k > 0, phase k 1 has been executed and the current sequent s k is provable. If at least one of the clauses R1 R4 in winning definition fails then Challenger wins. Suppose that all four clauses hold. We show that Challenger can choose a potent s k -native formula in such a way that every legal response s k+1 of Refuter is provable. Since s k is provable and condition R1 holds, Γ k is not closed under s k - native deduction. Hence there exist s k -native formulas ψ outside of Γ k such that sequent [Γ k ψ] is provable in SCF. Challenger should choose a formula z among such formulas ψ subject to an additional constraint: the proof P of sequent t = [Γ k z] should be minimal possible. Clearly t is not an axiom of SCF. Let R be the inference rule used to obtain t in P. Taking into account that conditions R2 R4 hold, R cannot be (Premise Inflation), ( E), ( I) or ( E). 13

14 We consider the remaining cases. In all those remaining cases formula z chosen by Challenger is potent. Case R = ( I), so that z has the form x y. Refuter guesses a sequent s k+1 = [Γ k+1 y] with Γ k+1 Γ k {x}. But the premise of (our application of) R is sequent Γ k, x y. So s k+1 is provable. Case R = (Said), so that z has the form p said x. Let be as in the definition of phase k. Refuter guesses a sequent s k+1 = [Γ k+1 y] where Γ k+1 consists of formulas native to {y} and Γ k+1. But the premise of R is sequent y. s k+1 is provable. Case R = (Implied) is similar to the previous one. Thus the question whether a given sequent s 0 is valid or refutable can be decided in alternating polynomial time. That concludes the proof of Theorem Hilbert-type calculus for primal infon logic The rest of this paper is devoted to primal infon logic. We introduce a Hilberttype calculus H for the logic. Let told with or without a subscript range over {implied, said}. A string π of the form q 1 told 1 q 2 told 2... q k told k is a quotation prefix; the length k of π may be zero. Let pref with or without a subscript range over quotation prefixes. If x is a formula q 1 told 1 q 2 told 2... q m told m y where y is a variable, conjunction or implication then every quotation prefix (q 1 told 1 q 2 told 2... q k told k ) with k m is a quotation prefix of x so that (q 1 told 1 q 2 told 2... q m told m ) is the maximal quotation prefix of x. We say that pref 1 is dominated by pref 2 and write pref 1 pref 2 if pref 1 is the result of replacing some (possibly none) occurrences of implied in pref 2 with said. Now we are ready to give the axioms and rules of H. Axioms: pref Inference rules: 14

15 (Pref S2I) (Pref E) (Pref I) (Pref E) (Pref I) pref 2 x pref 1 x pref (x y) pref x pref x pref y pref (x y) pref x pref (x y) pref y pref y pref (x y) where pref 1 pref 2 pref (x y) pref y A derivation of a formula ϕ from hypotheses Γ in calculus H is a sequence x 1,..., x n of formulas where x n = ϕ and every x i is an axiom, a hypothesis or the result of applying one of the rules to earlier members. H does not have the subformula property but it has a similar property. Q: How do you know that H does not have the subformula property? A: Here is a simple example. Hypotheses p said x, p said (x y), p said (y z) entail p said z. Indeed, the first two hypotheses entail formula p said y; that formula and the third hypotheses entail p said z. That derivation and indeed all derivations of p said z from the hypotheses contain formula p said y which is not a subformula of the hypotheses or the conclusion. Definition 5.1. The components of a formula z are defined by induction: z is a component of z, and if pref (x y) is a component of z or if pref (x y) is a component of z, then pref x and pref y are components of z. The components of a set of formulas are the components of the formulas in. Lemma 5.2 (Components). If x is a component of z then p told x is a component of p told z. Proof Induction on the definition of components. Further, a formula x is dominated by a formula y if x = pref 1 z and y = pref 2 z for some z and pref 1 pref 2. Definition 5.3 (Local). A formula x is local to a formula z if it is dominated by a component of z. Formula x is local to a set of formulas if it is local to a formula in. A prefix pref is local to if some formula pref y is local to. A derivation x 1,..., x n of ϕ from Γ in calculus H is local if every formula x i is local to set Γ {ϕ}. 15

16 Recall that SCP is the sequent calculus of 2.2 for primal infon logic. Theorem 5.4. For every sequent s = [Γ ϕ], the following claims are equivalent. 1. s is derivable in SCP. 2. There is a derivation of s in SCP where every formula is a subformula of s. 3. s is valid. 4. ϕ is derivable from Γ in H. 5. There is a local derivation of ϕ from Γ in H. Proof (1), (2) and (3) are equivalent by Theorem 3.2. (5) obviously implies (4). It suffices to prove that (4) implies (1) and (2) implies (5). (4) implies (1). By induction on a given derivation of ϕ from Γ in H, we prove Γ ϕ in SCP. If ϕ is an axiom pref of H, start with the axiom of SCP; repeatedly apply rules (Said) or (Implied) to obtain sequent (pref ); then repeatedly apply (Premise Inflation) to obtain sequent Γ pref. If ϕ is a hypothesis, use axiom (x2x) of SCP and then repeatedly apply (Premise Inflation). Otherwise several cases arise according to the last step in the given derivation of ϕ. We consider only the case when rule (Pref E) was applied at the last step; other cases are similar. By the induction hypothesis, Γ pref x and Γ pref(x y) are provable. By Lemma 2.3, it suffices to prove the sequent Γ, pref x, pref (x y) pref y. Start with an obviously provable sequent x, (x y) y and apply rules (Said), (Implied) to obtain sequent pref x, pref (x y) pref y; and then repeatedly apply (Premise Inflation). (2) implies (5). By induction on a given proof P of sequent s = [Γ ϕ] in SCP that uses only subformulas of s, we construct a local derivation of ϕ from Γ in H. The cases when s is an axiom are obvious. Otherwise several cases arise according to the last step in P. We consider here only two cases. ( E) Suppose that rule ( E) was applied in the last step of P. By the induction hypotheses, we have local derivations of x and of x y from Γ in H. It remains to apply rule (Pref ) with the empty pref. (Implied) Suppose that rule (Implied) was applied in the last step of P, so that Γ has the form (q said 1 ) (q implied 2 ) and ϕ has the form q implied ϕ 0. By the induction hypothesis, there is a local derivation D of ϕ 0 from 1 2 in H. Without loss of generality, D has the form 1, 2, Tail 16

17 so that first 1 formulas are listed, then 2 formulas are listed, and then the remaining tail formulas z are listed. Let D be q implied 1, q implied 2, q implied Tail. D is a derivation of ϕ from Γ = (q implied 1 ) (q implied 2 ) in H. Indeed, if a tail formula z of D is an axiom or is obtained from earlier members of D by means of a rule R then (q implied z) is an axiom or is obtained from the corresponding members of D by rule R. Furthermore, D is a local derivation. Indeed, if a tail formula x of D is dominated by a component y of 1 2 {ϕ 0 } then q implied x is dominated by q implied y which, by Lemma 5.2, is a component of Γ {ϕ}. Further, let D be (q said 1 ) followed by D : q said 1, q implied 1, q implied 2, q implied Tail. D is the desired local derivation of ϕ from Γ in H. First we check that D is a derivation. Since D is a derivation of ϕ from Γ ; it remains only to notice that q implied 1 is obtained from q said 1 by repeated applications of rule (Pref S2I). In fact, derivation D is local to sequent s. Indeed formulas in blocks (q said 1 ) and (q implied 2 ) are local because they are hypotheses. Formulas in block (q implied 1 ) are local as they are dominated by hypotheses q said 1. And, since D is local to Γ {ϕ}, every formula z in (q implied Tail) is dominated either by a component of ϕ or else by a component of Γ which is dominated by the corresponding component of Γ. Q: Do you really need local formulas in addition to the components of Γ {ϕ}? Maybe there is a derivation of ϕ from Γ that uses only the components whenever ϕ is derivable from Γ. A: Here is a counter-example to your conjecture. Let x, y, z be distinct infon variables, and let Γ be p implied x, p said (x y), p said (y z). Formula ϕ = (p implied z) is derivable from Γ but any derivation involves local formula (p implied y) that is not a component of Γ {ϕ}. Furthermore, if y is not a variable and y = (q said y ) then any derivation of ϕ from Γ involves a quotation prefix (p implied q said) that is not a prefix of any component. 6 Primal intuitionistic logic To logicians, primal intuitionistic logic may be of interest in its own right. In this connection, in 6.1, we specialize the relevant results above to the case of primal 17

18 intuitionistic logic. In 6.2 we construct a linear time algorithm for the multiple derivation problem (defined in 1) for primal intuitionistic logic. The algorithm will be generalized in the next section. For brevity primal intuitionistic logic will be called PC which is an allusion to Primal Constructive logic as intuitionistic logic is also known as constructive. 6.1 Syntax and semantics. PC formulas are built from variables and constant by means of conjunction and implication. A PC sequent calculus is obtained from the sequent calculus for primal infon logic by removing inference rules (Said) and (Implied): Axioms ( ) (x2x) x x Inference rules (Premise Inflation) Γ y Γ, x y ( E) Γ x y Γ x Γ x y Γ y ( I) Γ x Γ y Γ x y ( E) ( IW) (Trans) Γ x Γ y Γ x y Γ x y Γ y Γ x, Γ, x y Γ y Kripke structures for PC are triples (W,, C) subject to conditions K1 K5 in 2.1 and condition K6W in 2.2. Theorem 3.2 remains true. More exactly, for every PC sequent s the clauses 1 4 of Theorem 3.2 are equivalent. A Hilbert type calculus H PC for PC is a simplification of calculus H of the previous section: 18

19 Axiom: Inference rules ( e) ( i) ( e) ( i) x y x x y x y x y x y x y y x y y In the PC case, the components of a formula z are exactly the subformulas of z, and the local formulas of z are exactly the subformulas of z. Theorem 3.2 simplifies as follows. Theorem 6.1. For any PC sequent s = [Γ ϕ], the following are equivalent: (i) s is derivable in the PC sequent calculus, (ii) there is a derivation of s in the PC sequent calculus where every formula is a subformula of s, (iii) s is valid, (iv) ϕ is derivable from Γ in H PC, and (v) there is a derivation of ϕ from Γ in H PC where every formula is a subformulas of s. 6.2 Linear time theorem for primal constructive logic Theorem 6.2. There is a linear time algorithm for the multiple derivability problem for primal constructive logic. Given hypotheses Γ and queries Q, the algorithm determines which of the queries in Q follow from the hypotheses Γ. Proof Formulas local to (that is subformulas of) Γ Q will be simply called local. By Theorem 6.1, we may restrict attention to derivations where all formulas are local. The idea is to compute all local consequences of Γ. This is obviously sufficient. Without loss of generality we may assume that does not occur in Γ Q. It follows that does not occur in any local formula. Let n be the length of the input sequence Γ, Q. The key K(y) of a local formula y is the position of the first symbol of the first occurrence of y in the input sequence. Parse tree Run a parser on the input string producing a parse tree. The subtrees of the hypotheses and the queries hang directly under the root. Each node of the parse tree has a label that is or represents (according to the lexical analyzer) a variable or connective. The label length is O(log(n)) due to the lexical analysis phase of parsing. Extra tags mark hypotheses and queries; such a tag is not a part of the official label. 19

20 By induction define the locutions L(u) of nodes u. If u is a node with children u 1,..., u k, then L(u) = Label(u)(L(u 1 ),..., L(u k )). The locutions are being introduced for use in our analysis of the algorithm; the algorithm will not have the time to produce all the locutions. Each node u is associated with a particular occurrence of L(u) in the input string. The initial position of L(u) in the input string is the key K(u) of u. Nodes u and v are homonyms if L(u) = L(v). A node with the least key in its homonymy class is a homonymy original. A homonymy original u represents the locution L(u); its key is the key of the locution. We presume that the nodes u come with homonymy pointers H(u) initially set to nil. Homonymy originals Run the Cai-Paige algorithm [4] on the parse tree. The algorithm partitions the (keys of) nodes u into buckets B l according to the height l of u, that is the height (or depth) of the subtree rooted at u. Furthermore, every bucket is ordered according to the lexicographic order of locutions L(u). Note that two homonyms have the same height and thus belong to the same bucket. Homonyms are ordered according their keys. The Cai-Paige algorithm sets every homonymy pointer H(u) to the homonymy original of u. This wonderful algorithm runs in linear time. Preprocessing Create a table T of records indexed by the homonymy originals u such that L(u) is a formula. A record T (u) has five fields. One of them is the status field S(u) with values in the set {1, 2, 3}. The status field is dynamic; its value may change as our algorithm runs. All other fields are static; once created their values remain immutable. The status field S(u) determines the current status of the formula L(u). Formulas of status 1, 2, 3 will be called raw, pending, processed respectively. A raw formula has not been derived. A pending formula has been derived but remains a subject of some processing. A processed formula has been derived and processed. Initially every S(u) = 1. Traverse the parse tree setting the status S(u) to 2 if L(u) is tagged as a hypothesis. The static fields of the record T (u) are (, left), (, right), (, left) and (, right). Each entry in these fields is a sequence of (the keys of) nodes. To compute those sequences, traverse the parse tree in the depth-first manner. If the current node v is the left child of a node v with label then append H(v ) to the (, left) sequence of H(v) and move on. The cases where v is the right child of v or the label of v is or both are treated similarly. Processing Walk through the table T and process every pending formula L(u) in turn. When the processing of L(u) is finished do the following. Set S(u) = 3 indicating that L(u) is processed. If u is tagged as a query then output K(u). 20

21 The processing consists of firing, one after another, the inference rules applicable to L(u). Rule ( e) requires that L(u) has the form x y. If any of the formulas x, y is raw, make it pending. More exactly, let u 1, u 2 be the left and right children of u. If S(H(u i )) = 1, set S(H(u i )) = 2. Rule ( i) may be applied in two ways depending on whether we view L(u) as the left or right premise of the rule. The two cases call them the left and right cases are similar; we describe here only the left case where L(u) is the left conjunct x of a formula x y. For any such y that has been derived but x y is still raw, make x y pending. More exactly, for every v in the (, left) sequence of T (u) do the following. Note that L(v) = L(u) L(w) where w is the right child of v. If S(H(w)) > 1 and S(v) = 1, set S(v) = 2. Rule ( e) also may be applied in two ways depending on whether we view L(u) as the left or right premise of the rule. This time around the two cases call them the left and right cases are quite different. The left case is similar to the left case in the application of rule ( i) above. For every v in the (, left) sequence of T (u) do the following. Note that L(v) = L(u) L(w) where w is the right child of v. If S(v) > 1 and S(H(w)) = 1, set S(H(w)) = 2. The right case requires that L(u) has the form x y. If x has been derived and y is raw, make y pending. More exactly, let u 1, u 2 be the left and right children of u. If S(H(u 1 )) > 1 and S(H(u 2 )) = 1, set S(H(u 2 )) = 2. Rule ( i) has only one premise y = L(u). Any raw formula of the form x y becomes pending. More exactly, for every v in the (, right) sequence of T (u) do the following. Note that L(v) = L(w) L(u) where w is the homonymy original of the left child of v. If S(v) = 1, set S(v) = 2. This completes the algorithm. Proof of correctness Note that every pending formula becomes processed. Obviously only provable formulas become pending. To prove the correctness of the algorithm, it suffices to prove that every provable formula L(v) becomes pending. We do that by induction on the proof length of L(v). If L(v) is a hypothesis, it becomes pending when the table is formed. Otherwise several cases arise depending on the rule used at last step of the given proof of L(v). All cases are pretty obvious. We consider just one of those cases. Case ( i) where L(v) = x y and x, y have been derived earlier. By symmetry we may assume without loss of generality that x became pending first. Formula y is L(u) for some u. But then v occurs in the (,right) sequence of T (u). Accordingly L(v) becomes pending as a result of applying the right case of rule ( i) in the processing of L(u). Time complexity It remains to check that the algorithm is indeed linear time. Obviously the parse-tree and table stages are linear time. The only question is whether the processing stage is linear time. Instead we claim something 21

22 else. Note that the processing of a pending formula L(u) is done in a fixed finite number of phases. At each phase, we make some number k(u) of attempts to apply a particular inference rule viewing L(u) as one of the premises. Each attempt takes bounded time. The number of attempts is not bounded. It suffices to prove, however, that u k(u) = O(n). The proof is similar for all phases. Here we consider only the phase when we attempt to apply rule ( e) with L(u) as the left premise x. In this phase the number k(u) is the length of the (,left) sequence of u. But the (,left) sequences for distinct records are disjoint. And so u k(u) n. 7 Linear time theorem for primal infon logic We return to primal infon logic. We will be working with fragments of the Hilbert-type calculus H for primal infon logic. To recall H, see 5. In 4 we gave the obvious definition of the quotation depth of formulas. In the case of primal infon logic, another definition of quotation depth is more pertinent. Definition 7.1 (Primal quotation depth). The primal quotation depth of formulas is defined by induction: δ(x) = 0 if x is a variable. δ(p told x) = 1 + δ(x). δ(x y) = max{δ(x), δ(y)}. δ(x y) = δ(y) Further δ(γ) = max x Γ δ(x) for any set Γ of formulas. L. Recall Definition 1.1 of the multiple derivability problem MD(L) for a logic Theorem 7.2. For every natural number d, there is a linear time algorithm for the multiple derivability problem for primal infon logic restricted to formulas z with δ(z) d. Q: The definition of primal quotation depth is strange. The clause δ(x y) = δ(y) ignores δ(x). If foo is a quotation-free formula then δ((p said q said foo) foo) = 0. The formula involves quotation but its primal quotation depth is zero. A: Well, the strange definition makes the theorem stronger. Q: Does the additional strength matter? A: It does. For example it allows us to interpret authorization logic SecPAL [3] in the stratum of depth 2 of primal infon logics [10]. 22

23 7.1 Reduction to Main Lemma Definition 7.3 (Regular). A formula x is regular if it has no occurrences of. A derivation is regular if all its formulas are regular. R is the fragment of H obtained by removing from the language and removing the axioms from the calculus. Two formulas are equivalent if each of them entails the other in primal infon logic. Lemma For any formula z there is an equivalent formula z that is either an axiom or regular. 2. There is a linear time algorithm that, given a formula z, computes the equivalent formula z. 3. Any local derivation of a regular formula from regular hypotheses is regular. 4. There is a linear-time reduction of MD(H) to MD(R). Proof 1. Easy induction on z. We consider only the case z = x a in the induction step where a is an axiom. In that case z may be a. Indeed z a because a is an axiom, and a z by (Pref I). 2. Execute the regularization algorithm implicit in the proof of 1. One appropriate data structure is parse tree. Even if z is given as a string, in linear time you can construct the parse tree of z and then execute the regularization algorithm. 3. By the definition of local derivation. 4. Apply the regularization algorithm to the hypotheses and queries. If a modified hypothesis is an axiom, remove it. If a modified query is an axiom, mark it derivable and remove it. Lemma 7.5. Theorem 5.4 remains true if sequent s is assumed to be regular and if H is replaced with R. Proof The proof of Theorem 5.4 needs only two minor modifications, in fact simplifications. In the proof that (4) implies (1), ignore the case where ϕ is an axiom of H. Similarly, in the proof that (2) implies (5), in the proof that D is a derivation, ignore the case where the tail formula z is an axiom of H. Lemma For any inference rule in R, the primal quotation depth of the conclusion is bounded by the primal quotation depth of the premise(s). 2. If Γ entails ϕ in R then δ(ϕ) δ(γ). 3. The restriction R d of R to formulas z with δ(z) d is a calculus in its own right; all inference rules of R produce formulas in R d when applied to formulas in R d. Proof Claim 1 is obvious but note that the unusual clause δ(x y) = δ(y) in the definition of primal quotation depth is used in the case of rule (Pref I). Claims 2 and 3 are obvious as well. 23

24 Main Lemma 7.7. For every natural number d, there is a linear time algorithm for the multiple derivation problem MD(R d ) for R d. Clearly Theorem 7.2 follows from Main Lemma. We prove Main Lemma in the next subsection. 7.2 Proof of Main Lemma Given a positive integer d, we construct a linear time algorithm for MD(L d ) by modifying the linear-time algorithm of 5. Recall that told with or without a subscript ranges over {said, implied}, and pref with or without a subscript ranges over quotation prefixes q 1 told 1... q k told k where k is the length of pref. We say that a quotation prefix is relevant if its length is d. Parsing Parse the input as in 6.2 except that now we have additional labels p told. There is a problem, however. In 6.2, every local formula was the locution L(u) of some node u of the parse tree. Now it is not necessarily the case. To remedy the situation, we graft extra nodes into the parse tree. This is not the optimal remedy but it simplifies the exposition. By induction on nodes u, define pref(u). If u is the root, then pref(u) is empty. Suppose that v is the parent of u. If the label of v is of the form p told then pref(u) is pref(v) appended with the label of v; otherwise pref(u) = pref(v). Let C(u) be the formula pref(u) L(u). It is easy to check that every C(u) is a component and that every component is C(u) for some u. Call node u relevant if (i) u is not the root of the parse tree so that L(u) is a formula and (ii) pref(u) is relevant. The fan F (pref) of a prefix pref is a tree of prefixes. It contains all prefixes pref pref. Further, it contains a prefix q 1 told 1... q i told i of length i if it contains a prefix q 1 told 1... q i told i q i+1 told i+1 of length i + 1 which is a parent of q 1 told 1... q i told i. Every node in F (pref) has at most two children and at most one parent. If d = 2, then F (pref) contains at most 7 nodes. Now turn F (pref) upside down, and let F (pref) be the result. Normally our trees go downward so that the root is at the top. F (pref) grows upward. Traverse the parse tree in the depth-first manner and graft a fresh copy F (u) of F (pref(u)) at every relevant node u. There is a one-to-one correspondence ξ : F (u) F (pref(u)). If v F (u) and ξ(v) is a prefix q 1 told 1... q i told i of length i > 0 then the label of v is q i told i and the key of v is the pair (Key(u), told 1... told i ) The keys are ordered lexicographically. We do not distinguish between Key(u) and the pair (Key(u), s) where string s is empty. Every node u of the resulting parse structure has at most three parents, and the nodes u form a tree. Remark 7.8. For every relevant original node u, the formula pref(u) L(u) is a component of Γ Q. If pref pref(u) then pref L(u) is local. Every local 24

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